Question

very important question guys
answer me​

Answer 1

Answer:

Step-by-step explanation:

the minimum value of

y= 5x^2 -4x +3

If we take y=0 , then x becomes non real as 'D' or (b^2 -4ac) becomes < 0

If we take y = for any negative ( -ve )value , x becomes non real

=> y > 0 ,

y = 5x^2 -4x +3

Here, in the above function we need the minimum positive value of the function

=> 0 = 5x^2 -4x +(3-y)

x = {-b +,- √(b^2 - 4ac)} / 2a

=> x = { 4 +,- √(16 - 4*5*(3-y) ) } /10

If we make D= 0 , we get the minimum value of the function.

For making D= 0

4ac = 16

=> 20(3-y) = 16

=> 3-y = 16/20= 4/5

=> y = 3 - 4/5

=> y = 11/5

Minimum value of the function= 11/5

Answer 2

$\huge\colorbox{red}{Answer:-}$

The minimum value of

y= 5x^2 - 4x +3

If we take y=0, then x becomes non real as 'D' or (b^2-4ac) becomes < 0

If we take y = for any negative (-ve )value, x becomes non real

y = 5x^2 - 4x +3

Here, in the above function we need the minimum positive value of the function

=> 0 = 5x^2 - 4x +(3-y)

x=(-b+ √(b^2-4ac)] / 2a

=> x= [4+, √(16-4 × 5 × (3-y))]/10

If we make D= 0, we get the minimum value of the function.

For making D=0

4ac = 16

=> 20(3-y) = 16

=> 3-y = 16/20= 4/5

=> y = 3 - 4/5

=> y = 11/5

Minimum value of the function= 11/5.